text:
Could higher aspect ratio actually reduce gliding efficiency of a smaller glider?
I have been building 1 meter wingspan 100g class free flight mono wing gliders with a chord of around 10 cm and a flying speed of approximately 3 meters per second.  So far glide ratios of around 10:1 have been achieved.  The wing thickness is 1.2 centimeters.  The airfoil
has maximum thickness at 3.5 cm from leading edge.  The bottom is flat, but only covered 6 cm from leading edge, leaving an undercambered area the remaining 4 cm to the trailing edge (this design grew out of thin undercambered wings).
So, is there a maximum aspect ratio  (possibly based on Reynolds number calculations and/or "pressure leakage" due to the small scale or low speed)?  Or do we continue to
go for as high an aspect ratio as structurally possible.
Reynolds number for airfoils = Vc/v where V = 3 m/s, chord = 0.1 meter, kinematic
viscosity at sea level = 1.460 × 10 minus 5th power = 20,548 for my model.
A response by Peter Kampf June 22, 2018 to "Is it best to use a thick airfoil or an undercambered airfoil for for slow speed flight" mentioned the Daedelus airfoils DAE 11,
DAE 21, DAE 31, so it seems to be on the right track.
I am getting ready to start a build, would there be an optimal aspect ratio for this
new aircraft?

The rule that gliding performance increases with aspect ratio is under assumption that the wing area is the same. But you have fixed span and trying to vary the area, so it does not apply.

@JanHudec I submit area can vary somewhat because the lift/drag ratio is being considered.  There is a weight constraint, so your point has some validity.

With again thanks to other writers who have inputted valuable information, it has been found that excessive aspect ratio, by way of Reynolds Number, can effect gliding efficiency
of a smaller, slower glider negatively as follows:
Reynolds number = Vc/v.  Slower speed and shorter chord reduce Reynolds number. Kinematic viscosity of air v does not change.
Using Airfoil Tools, Coefficient of Lift/Coefficient of Drag at various Alpha was evaluated using a range of Reynolds Numbers from 50,000 to 1,000,000 for the Bleriot Eiffel 428 (thin undercambered)  and the Wortmann FX 60 126 (thicker, less under camber).
It was found that Clift/Cdrag is significantly greater, at higher Reynolds number, particularly in a "sweet spot" around 6 degrees AOA.  In this region, drag seems to "plateau", while Clift continues to rise.
At very low Reynolds Numbers, the Eiffel 428 showed a slightly better Clift/Cdrag ratio, but in ALL cases higher Reynolds number improved Clift/Cdrag ratio up to around 500,000 (will vary depending on air foil type), after which the polars seemed much more consistant.

First do the essential numbers:
Wing span is 100 cm, chord is 10 cm. So your wing area is 0.1 m². I assume standard atmospheric density, assuming you live close to sea level. This makes your lift coefficient with 100 g of mass and 3 m/s
$$c_L=\frac{2\cdot m\cdot g}{\rho\cdot v²\cdot S}=\frac{1.96}{1.1025}=1.78$$
while the Reynolds number is only 20,534. This doesn't fit together – not even Airfoiltools goes that low. Your speed is most likely higher (the other possibility is flight somewhere in Siberia, which I discount for now).
More than a $c_L$ of 1.2 at those Reynolds numbers is unlikely. This would be reached at 3.65 m/s and Re = 25,000. Your best glide will be at this lift coefficient with an aspect ratio of 10 and a zero-lift drag of 0.05 (let's for now use a classical model airfoil). If you follow the Airfoiltools results, even at twice the Reynolds number the E387 has a large, laminar upper side separation bubble that requires the use of a turbulator. At Re = 25,000 the maximum lift coefficient is well below 1.2 and drag rises from 0.03 at $c_L$ = 0.6 to 0.07 at $c_L$ = 1.1. I smoothed the Airfoiltools coordinates for the E387, increased camber and thickness a bit to come as close to the airfoil you describe and placed a forced transition at 30% of chord on the upper side. Still, its high drag would give a best gliding lift coefficient of around 0.9 at the Reynolds number of 29,000 that this lower lift coefficient would suggest.
You are most likely better off with a bird-like airfoil of less thickness and more camber. With what you have the aspect ratio of 10 looks too high, which is confirmed by the lower aspect ratio of all smaller birds. Using $c_L$ = 0.9 and $c_{D0}$ = 0.035, the aspect ratio to make this point that of best glide is only
$$AR=\frac{c_L^2}{c_{D0}\cdot\pi\cdot\epsilon}=\frac{0.81}{0.1045}=7.75$$
So my answer to your question is Yes.

Re "which is confirmed by the lower aspect ratio of all smaller birds" -- consider swifts?

@quietflyer yes, and seagulls. They fly a lot faster (hence the name) which shifts their Reynolds numbers up considerably. This in turn increases the optimum aspect ratio, too.

From your description, your glider is squarely in the laminar flow regime, barring any unintended turbulator due to manufacturing.
From an aerodynamic perspective, you'll be trading off skin friction drag with induced drag. Induced drag is characterized by: $C_{D_i}=\frac{C_L^2}{\pi e A}$, whereas skin friction drag is: $C_{D_f} \approx 2\frac{1.328}{\sqrt{Re_c}}$ (source: Anderson, Fundamentals of Aerodynamics). $A$ is aspect ratio, $S$ is wing area, $Re_c=\frac{\rho V c}{\mu}$ is the characteristic Reynolds number of the wing.
Assume the simplest case of a rectangular wing, characteristic chord is $c=\frac{S}{b}=\sqrt{\frac{S}{A}}$. You can use a spreadsheet to find out where an increase in aspect ratio begins to increase your drag, assuming surface area remains the same.
The structural perspective is harder to nail down. I would imagine that the structural members are minimum gauge (i.e. sized not for strength but for manufacturability) even for the smaller aspect ratios. But as you increase in aspect ratio, there will be a weight impact.

"barring any unintended turbulator due to manufacturing", I had heard balsa gliders (and frisbees) actually benefit (on their scale) from a little surface roughness.  There is something about "tripping" the flow or "energizing" it that helps with efficient lift creation (golf ball dimples too).

“where an increase in aspect ratio begins to increase your drag, assuming surface area remains the same”—the question clearly states span is 1 m, so the area is the variable here. Slightly different calculations are called for!


text:
Why did the DC-3 have a much higher cruise speed than the HP.42, despite their similar weight and power?
A review of the classic DC-3 and the Handley Page HP.42 yielded a surprisingly close match in empty and max gross take off weight at around 18,000 and 28,000 lbs for the HP.42 and 16,000 lbs and 25,000 for the DC-3.  The HP.42 cruised at 87 knots on around 4x500 hp, and the DC-3 at 180 knots on 2x1200 hp.  The HP.42 was a lower aspect biplane, the DC-3 a graceful mono plane, the epitome of aviation advances for that era but...
Questions:  How much of the DC-3's speed increase was the result of improved engine/prop efficiency?  How much from the DC-3's speed increase came from its beautiful Prandtl inspired wing?  How much from retracting its gear?  How much closer would the HP be with those improvements, while remaining a biplane (perhaps with improved struts)?
I do not expect an exact answer, I am more interested to see how the numbers were crunched and a short explanation why.  Thanks.

Struts every which way, uncowled engines, wheels with fenders hanging out, wires, bits sticking out here, bits sticking out there, corrugated skins, fabric wings and tail... it's a mess; a flying Truss Bridge.  There's just too much to try to quantify it like that.
You could probably take a DC-3 and start adding all those things one at a time, and each item would knock off 10-20 kt, and eventually you'd end up down at 87 kt, but the slower you get the smaller each accumulative penalty would be.

True, oh John.  I am going for rough percentages, particularly the engines.  I am not sure if the HP.42 had variable pitch props.  The transition from 4 to 2 engines echoes the trends today.  Are we too infatuated with low drag wings?  But the Prandtls on the HP or (please!) slats on the DC would be major improvements.  Stall on the DC was controlled by aileron into the stall (hoping to stall the other wing) .  Slats seem so much better.

I think that implies that the performance difference was nearly all down to reduced form drag, and not improved prop efficiency.

The props are fixed wood ones, but if they are optimized for cruising, the difference between constant speed and fixed is not that great all else being the same.  If optimized for climb then they are underpitched for cruise, and there is penalty at cruise of maybe 10-20 kt.  If it had "climb" propellers and did 87kt, changing the props to ones with more pitch, "cruise" propellers, might add 10-20 kt, maybe.

In the case of biplane and a monoplane where all the wings involved have the same aspect ratio, the extra wingtips of the biplane should cause no loss of efficiency.  On the other hand, interference between the top and bottom wings of the biplane will cause some loss of efficiency.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Is the highly optimized high subsonic cruise airfoil less safe for higher AOA flight?
An examination of the 737 airliner wing profile and that of a blue whale leads to the question of whether the high Reynolds number subsonic cruising airfoil and the low speed, high AOA airfoil are compatible.
Supercritical airfoils are designed for efficient cruising flight, relying on bottom lift and minimizing acceleration of air over the top to avoid creating drag producing transsonic shock waves.
The blue whale is designed to cruise efficiently over thousands of miles, through a viscous, non-compressible fluid (water). Their shape includes a rounded, sloping lower fore and a flatter top, very similar to a supercritical airfoil!  Applying thrust to this shape would provide "lift" simply by directing water "down" (or pushing the whale "up").
Reynolds Number = Velocity×Chord/Kinematic Viscosity
Reynolds Number 20th century Type VII U Boat (submarine):  speed: 4 meters/second
                  chord length:  70 meters. Kinematic viscosity (sea water): 1.04 × 10e-6
      4 × 70/1.04 × 10e-6 = 269 million

Reynolds number Blue Whale:  speed: 5 meters/second, chord length: 30 meters
                       Kinematic viscosity (sea water):  1.04 × 10e-6
      5 × 30/1.04 × 10e-6  = 144 million

Reynolds number Supercritical Airfoil (approximate):  speed 300 meters/second,
Chord length:   4 meters    Kinematic Viscosity (air):    1.46 × 10e-5
    300 × 4/1.46 × 10e-5 =     82 million

Reynolds Number Cessna 172 Airfoil :  speed 30 meters/second
Chord length:  2 meters        Kinematic Viscosity (air): 1.46 × 10e-5
   30 × 2/1.46 × 10e-5 =         4 million

Reynolds Number Albatross (bird) airfoil: speed 10 meters/second
Chord length: 0.2 meters       Kinematic Viscosity (air): 1.46 × 10e-5
  10 x 0.2/1.46 x 10e-5 =        140,000

We can see the supercritical wing is more in the range of the blue whale!
When pitched up to a higher AOA, this shape would be a very poor wing in the classic Bernoulli sense, as the significantly contributing top lift would be lost at a lower AOA than a top rounded airfoil, behaving much more like a flat plate (see polars).
There are accounts of airline pilots extending their slats at higher speeds for better performance (and getting into trouble for doing so).
So, for modern designers, would it not be important to consider a more robust low speed system that could be left on (in varying degrees) at higher speeds (V maximum flaps and slats extended airspeed) to help avoid stalling?
This could create a wider safety margin for the climb-out and landing phases of flight. Hard earned efficiency gains in cruising flight (from the supercritical wing) would be kept, when they are fully retracted, but only at safe speed and AOA conditions (for example, above 10,000 feet or at cruising flight level).

Blue whales and their like, minke and sei, are amoung the fastest of the whales. They could not be caught by sailing ships.  If only buoyant, they would be fully symmetrical (like blimps). Look again at their profile, and please give me a Reynolds number of a 35 meter blue whale swimming through water at 20 knots.  My guess is they are both bouyant and partially lifting (like some modern airships)

I think the whole "whales" thing is clouding your question and will get people off-topic. Whales use other "lift devices" than just the fins (like air in the lungs) so I don't think you can do an apples-for-apples comparison here on the other lift devices. Plus whales don't travel at trans-sonic speeds, even if they are fast.

@ymb1 The underbelly fairing of the Pilatus PC-24 in another question got me thinking.

If you do not want to use Mathjax for your equations, at least please use scientific notation for powers of ten, like this: 2.5e-5 = $2.5 \cdot 10^{-5}$, the way it is now is very hard to read.

OK, and we'll lower the whale cruising speed a bit too.

When calculating the Reynolds numbers, you should use a length that is representative of the foil. If you don't use the fuselage length for the C172 why do you use the length of the body and hull for the whale and submarine, instead of the chord of the caudal fin and dive planes? Not that I believe those things add anything to the question; it would be much improved without all the off-topic stories.

@AEhere because on the submarine IS the foil.  This may have applications in lifting fuselage design, as well as understanding the dangers of supercrits at lower Reynolds Numbers.  As far as "off topic stories", won't you let an old man tell his (relevant) tale?

No, it is not, submarines are trimmed for neutral bouyancy using ballast. Their mission requirements need them to remain at a set depth regardless of speed, and using the sub as a foil would be 1) a massive headache for the skipper and 2) woefully inefficient, given their AR of 0.09 or so.
Regarding the story interlude, it is welcome when on topic and furthers the question, but this is a Q&A site, not a forum, and in this case comparing a M=0.85 foil to a whale flipper is absurd.
Otherwise, you could always try chat.

@AEhere  No, we are talking about the entire whale.  And how about combination lifting and buoyant airships.  Or maybe airplanes should be all buoyant and no lift because it's such a pain!  C'mon AEhere.  We're talking about lifting bodies and wings, and application of the form. I little lift from the fuse is better than none, and using a supercrit for a wing requires knowledge of its virtues and limitations.

PS a wartime submarine design should also include lifting abilities should ballast systems fail. Look again at that U boat.

I was going to let this argument die like it deserves, but I decided to do the math: for a Type VIIC to produce half its own weight through slender body lift, assuming a 10 degree AoA and a lift curve slope of 0.02 (in rad), it would need to reach 32 m/s, or 87 knots. Few surface vessels today can attain such speeds, because slender body lift is inherently inefficient. So no, even without getting into the noise generated by such a lifting mechanism, it is a no-go for submarine design. The only cases I have come across towed ROVs that need to counter the vertical pull of their cable.

@AEhere No, you use a combination of buoyancy and lift. A damaged submarine will still have some buoyancy.  You need to relax.  Now, back to whales.  "Right" whales floated when they died. Some sank, and could not be recovered. See?

And BTW, if you take a gander at the polars, 6 degrees AOA for L/D is more like it (in air).  As a researcher, I would consider keeping it there and increasing lift by increasing camber (if that could be variably controlled) and power.  I would dig out a Storch and start there.

That's why I calculated for half. You can redo this calculation for your whale and tweak it to your heart's content, but you will find you can't get it to produce enough lift as a slender body. I would suspect that whale flotation post mortem depends on what organs have been pierced, but the fact that some float proves that they don't need forward speed for it. Anyway, I'm out unless you can produce numbers to back up your claims.

@AEhere They actually had to pump air into them.  But good discourse and thanks for the numbers.  I'm just trying to get them to leave those slats down a bit longer, and maybe set the stick shaker threshold a little lower (AOA).  Drag increases rapidly over 6 degrees for supercrits.  Over and out.

The short answer, whales aside, is yes.  Such wings, called "supercritical"  because the critical Mach# is elevated, tend to suffer from leading edge stall that makes the aircraft's natural stall behaviour dangerous.  These airplanes typically require stick pushers (as @Peter Kampf mentions in this answer, the 737, even the later ones that got redesigned wings, doesn't really have a "pure" supercritical wing and I believe that's why they don't require stick pushers).
The Challenger business jets and the CRJ line of Regionals all have supercritical wing  profiles.  The profile promotes a pre-stall flow separation and immediate reattachment at the LE, called a Laminar Bubble, which would form at some AOA prior to stall.

This had an effect similar to having a stall strip or ice shape, with the bubble forming a trip wire so to speak.  It causes the wing to stall at the leading edge instead of advancing from the trailing edge, which meant the entire wing just let go all at once with no buffet or other warning, with a tendency to progress right into a deep stall.  The Challengers and RJs have stick pushers because of this; they must never be allowed to stall naturally, because it is probably going to be unrecoverable.  Bombardier did a lot of testing with various aerodynamic band-aids like vortex generators and vortilons and such, and they didn't help.
The CRJs larger than the 200 have LE slats and are just about unstallable slats out, but they also have the same LE stall slats retracted and still require stick pushers.  Flying with slats extended at high speed is out of the question. The drag they produce is massive, as you discover when you first land one after flying a hard leading edge CRJ200. Incorporating a fixed slot in the leading edge could be done but they are pretty draggy and would negate the benefit you're trying for in the first place, and I'm sure there are other mach related issues with a fixed slot.

Good explanation. The other side of the story is the wing sweep for these types of aircraft, which contributes to pitch up post stall.

Thks. I have a chum who's experienced a natural stall on a CRJ200, when extruded LE sealant (the a/c was released before it had fully cured) tripped the flow on one wing PRIOR to pusher when he was doing the shaker/pusher test.  The shaker started going then suddenly it kind of quarter snap rolled.  I think what saved them was instead of going into a spin or deep stall it just fell away while knife edge and ended up in a dive.

@John K certainly would keep the supercritical at cruise.  It may be helpful to be able to extend slats at a higher airspeed than currently available to give a pilot more safety margin while manuvering, retracting them for cruise under safer conditions.

@RobertDiGiovanni that's just not really a problem on an airliner.   As it is you usually have slats out to the initial setting not that much below 200 kt on an arrival (on the RJ's the first slat/flap extension you do passing below 190kt). For maneuvering at high G however, the F-86 had automatic slats for exactly that purpose.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What bank angle creates the least energy loss in an emergency engine out 180 degree turn?
A situation we all hope to never have:  engine quits at a low distance AGL.
What bank angle will produce the least altitude loss for a 180 degree turn?
Is this common for all aircraft?
Readings have shown the elevator/wing to be more efficient at turning the aircraft
than the rudder alone.
It would seem better not to try to hold altitude on the turn, but one still must avoid stalling
the wing.  Airspeed is also a factor in gliding efficiency.
What is the best way to turn 180 degrees with least loss of altitude?
Some good references provided.  Curious about rudder technique in a steeper bank. At 45 degrees, rudder is pitching nose down some, but should be there to "follow" the wing around the turn (lowest drag).  Some elevator, some rudder to swing the tail through the yaw plane, giving up some altitude, so as not to load the wing as much (like an emergency descent)?

Why do we think this doesn't vary hugely between different aircraft?

It's a lot steeper than most people think.  Somewhere around 45 deg or more.  When I'm thermalling in a glider at minimum sink, I'll bank to around 45-50 deg to get the smallest turn radius without the sink rate going up significantly.  On your airplane, you have to figure it out by experimentation.

Even if this does vary hugely between different aircraft, I think this is still a good question that admits a good answer—that answer would just have to point out that it varies between different aircraft, and how and why. Conversely, maybe it really doesn't vary much between different aircraft.

Turn radius being independent of weight suggests similarity, although the fuse design could be extremely variable.

You should consider the plane AND the wind AND the IAS when deciding on an angle for an emergency 180. Some planes may stall at 45 degrees, for example, depending on the speed. You may want a steeper turn if you need to travel some distance after a turn into a headwind. As a side note, if you need to be aligned with a runway after the turn, consider the crosswind direction when deciding which way to turn. Another side note: Only practice this at altitude or on a simulator. People die regularly practicing emergency 180s at low altitudes.
This question is not a duplicate and should be open.

@xpda agree with last statement, as we are discussing technique.  Yes, practice at altitude and have landing sites straight ahead if possible. Bank angle doesn't stall the plane, yanking the elevator to try to hold altitude does.  This is why I was hoping for input using emergency descent technique with lower Gs (less loading to avoid stall).  Although theoretically possible, a near stall turn would not be on the option list.

Asked and answered, since a any plane without a functioning engine is basicly a glider:
What is the optimal bank angle to accomplish a given turn in a glide?
Assuming the question refers to an engine failure after takeoff, there is a lot more to the stunt than just turning with the least amount of altitude loss. In any case, before attempting "the impossible turn" you must be above the safe altitude for the maneuver. If you have not yet reached safe altitude, you make an emergency landing in the front sector. The safe altitude varies heavily with different plane makes and models, but for example a C 172 needs about 500ft and rather aggressive (yet skilfull) handling according to tests made for AOPA Pilot 2002 july article.
To make an efficient 45 degree turn while maintaining a steady speed is not exctly a walk in the park if you have just lost an engine. Attempts to return to the runway after an engine failure during initial climb have a strong tendency to end in tragedy.

Welcome to Stack Exchange! I'd like to point out that this question is quite specific as to what's considered the "best" bank angle ("least altitude loss for a 180 degree turn"), while the other question is more vague about it ("the greatest turn rate with the smallest loss in altitude"—without specifying how to handle the trade-off between the two contrary objectives).

This is true. I made the bold assumption that the reason to make a 180 after engine loss is to get back to the runway, Why else would you waste precious energy in a life threatening situation? To achieve this goal, you should use the best compromise between rate of turn and loss of energy. Less than 45 deg. bank angle, you drift further away, more than 45, you waste energy having to pull harder and trading altitude for speed not to stall as the load factor increases rapidly, and so does your stall speed. Plus it gets increasingly hard to make fully coordinated turns at higher bank angles.

The simplest answer to the question "What bank angle creates the least energy loss in an emergency engine out 180 degree turn?" would be "more than zero, less than one degrees". This way there would be virtually no energy lost due to the heading change, but it would take ages to make the turn :)

@Jpe61 The question is not asking about the altitude lost "due to the turn" but "during the turn". If your turn took an hour, then lots of altitude would be lost during that hour - much more than for a quicker turn with a higher bank angle.

The headline asks one thing, the body kinda the same but still different. Hence the smiley at the end of my previous comment.

We went to an FAA seminar on the emergency turn back to the airport, and 45 degrees was given as the best bank angle after testing by an experienced pilot. Also, taking off and climbing while turning to 45 degrees away from the runway, so that if one had to turn back, the 180 would get you better lined up with the end of the runway, vs taking off, doing a 180, then a further 45, and then 45 the other way to be lined up on the runway. (Sitting here waving my hands in the air picturing this.)

What's interesting is that in soaring training, we do simulated rope breaks at afew hundred feet.  Students are shocked to discover that not only that there is no problem making it around from as low as 300 ft, the real challenge is ** not overrunning the freaking landing zone ** running downwind if there is any significant wind taking off.


text:
Would a 50 mph aircraft holding a ground pattern in a 20 mph cross wind risk stalling by abruptly rolling away from a crosswind?
This is another play on the "down wind turn myth".
We all know a plane flying in a "uniformly moving airmass" is not affected by wind direction and speed.
But a plane following a ground track must compensate for wind direction and speed.
In this scenario:
Pilot Bob could not wait to fly his ultralight and went out to practice take-offs and landings on runway 270 with a 20 knot westerly wind from 270.  In the pattern, flying heading 180 on the crosswind leg, a heavy crab into the wind had to be applied to hold the ground track.
Instead of adding power, Bob increased AOA to hold altitude (BAD MOVE).
Flying near stall, the pilot abruptly rolled and yawed the aircraft away from the crosswind and began to turn into the 090 "downwind" leg.
Could this maneuver significantly change the AOA of the wing, resulting in a stall?

Comments are not for extended discussion; this conversation has been moved to chat.

Your question is flawed because it is creating a link between two different things.  Let's deal with them one at a time:
In the title question you ask whether a plane holding a crosswind correction in a landing pattern risks stalling by rolling abruptly away from a crosswind.  Ignoring for a minute that doing anything "abruptly" imposes some additional level of risk, (depending of course on the severity) the answer is no.  Turning into, or away from the wind in a "uniformly moving airmass" does not alter the airspeed of the aircraft.  You said as much yourself.  When ready for downwind you simply turn to a new heading for that leg as you would normally.  (in this case 090 because you will now have a tailwind.)
The very concept of "wind" only matters when you consider the surface of the earth.  The airplane doesn't recognize or care what the ground is doing, only the airmass matters.
Now to the second part; would flying near stall speed and then abruptly rolling and yawing the aircraft risk a stall?  Yes, absolutely yes!  As well as a spin...  Those are the main two ingredients of a spin, and every pilot is taught to avoid them, especially down low.  Statistically this is probably one of the main GA pilot killers.
So, let's further unlink the two concepts.  You seem to be espousing an unbalanced slip approach to maintaining heading and ground track in the crosswind turn, furthermore, doing it right around stall speed!  (Sorry, that wasn't you, it was pilot Bob...) This is improper and unsafe technique.  You should not attempt to maintain aircraft heading of 180 on the crosswind, you should fly 190, or 195, or whatever heading will allow you to maintain a consistent ground track.  You should do it wings level, in balanced flight, at a safe and appropriate airspeed for your aircraft in the pattern.  The same airspeed you fly when the airmass is not moving relative to the ground.  This wind correction technique is called "crabbing".
Bottom line, it isn't the wind creating any additional risk, it is pilot Bob's improper technique.

Comments are not for extended discussion; this conversation has been moved to chat.

Yes. Any abrupt maneuvering near stall speed in any conditions increases the risk of stall. Add yaw (flying an uncoordinated turn), and you further increase the risk of entering a spin.
As for the effects of the wind in this and similar situations: If the wind is steady, it has no effect. John K made an excellent remark in comments about turning close to the ground (in a glider), where strong wind gradient may cause control issues. However, standard landing pattern height is 1000 ft (300 m), and at this height wind gradient for 20 kts (~10 m/s) is only about 1% over a 33 ft (10 m) of height (the table only goes to 150 meters (500 ft) but gradient diminishes as altitude increases further). For a plane with 10  meter wing span in a 30 degree bank turn, the difference in wind velocity over wingtips would be less than 0,2 knots. That order of magnitude is unnoticeable. In the particular case in question, the whole paragraph above is irrelevant, as Bob has already stalled his plane and is heading for the ground. Poor Bob.
Below is a link to an article which very profoundly explains the physics of the downwind turn, the reasons it is not dangerous for, as well as the reasons it is dangerous for (attention, a bee is involved in the examples):
Australian flying: Dragons of the Downwind Turn

Fact 1 - A downwind turn is not dangerous - aerodynamically
Once the aircraft leaves the ground and is flying in a steady wind, its aerodynamics relate only to its movement through the air in which it is flying. Its geographical position is dependent on the wind’s speed and direction – but that is a navigational matter – not an aerodynamic one.

(should a request appear, I'll quote the article more here)
And since this subject has proven to be complicated enough that I have failed to communicate the physics of the downwind turn to several fellow aviators on two previous attempts, a little humour on the subject might lighten up the atmosphere and make the subject easier to grasp:
Flying Magazine: The Last Word on Downwind Turns, Really.

Good answer.  Now, in comparison with a rising wing having a lower AOA than a descending wing, will rolling away from a crosswind change the AOA of the wing???

Any time you are rolling the upgoing wing temporarily has a higher AoA than the downgoing wing.  That is what produces the roll.  And in a level turn, AoA overall must be increased slightly to make up for the reduced vertical component of lift.  (assuming a constant airspeed)

The airplane doesn't know it's in a crosswind.  The concept of a downwind turn stall is basically a reduced performance windshear scenario, but for that to work you would have to turn instantly.  Normal turns take too long for any inertial effects to work.

@MichaelHall one would typically produce roll with ailerons, and in that case the upgoing wing has lower AoA, and downgoing wing has higher AoA than in straight and level flight. One can roll the aircraft with rudder too, but that's not the usual way.

@Jpe61, OK, I meant lift, but maybe I should have said camber.  In any case I'm not understanding Robert's confusion...

Same here...  I've learned years ago this is a subject that is very hard to explain to some peers. Not all get it even in the long run. It kind of does not matter, as the myth keep people on their toes, so it guides them on the safer side of flying.

It can be tough to shed the the earth-centric paradigm.

No, you can't turning around a point.

@John K you have come the closest to understanding this with "normal turns take too long for inertial effects...".  Great, that was my bumblebee/747 concept.  Now, let's forget the turn and concentrate on the roll (useing sideslip technique), what happens to wing AOA if you roll away from crosswind?

Inertial effects of what? The wind?

@Jpe61 inertia of the aircraft being turned faster than it "speeds up" on its new heading.  That was bumblebee/747 (not bumblebee in 747). So light a cigarette, and walk slowly around a small circle in a wind, what does the smoke do?  Now get in your car and drive 60 mph around a circle 1/2 mile in radius.  That is what was meant by "scale".  But this question is about "roll".  But good knowledgeable inputs anyways, I'm going to test at a safe altitude.

So, with cigarette, stick your arms out and roll into crosswind, then roll away.  Will this change AOA of your "wings"!

Please do make a 360 midair circle in steady wind with some kind  of smoke apparatus (I refuse to advocate smoking) and find out, that at any steady turn rate and speed, you'll end up exactly where the smokevane started. It's totally different from what happens in your example of doing this on the ground.

As Jpe61 said, if you drive around in a circle on a dirt road (not a smoker), you will come back to your tire tracks. You will not come back to your dust trail unless it’s a perfectly calm day (hyperbole). The dust will drift in the wind. If you do a commercial standard steep turn in an airplane, you will run into your own wake turbulence. Both you and your wake are drifting in the wind.

There is an issue that affects gliders that bank close to the ground in a strong crosswind. The wind gradient can cause the glider's high wing to overbank the craft but the risk isn't a stall so much as just rolling into the ground from lack of roll power to overcome it in extreme cases.

This is an excellent remark @JohnK , I incorporated it into my answer after finding some data to show wind gradient at standard LDG pattern height. At lower altitudes, it seems obvious this might cause concern to gliders.

You seem to think that pilot Bob has to have a sideslip at some point in the pattern, because he has to make 90 degree turns in the air, and somehow also has to follow the groundtrack with 90 degree turns.
However, that's not the case (at least not if Bob is anything resembling a good pilot).
If he wants to follow a groundtrack with 90 degree angles, he has to adjust his heading, and the path he flies in the air looks something like this: (the wind is coming from the right)
     <-<-<-<-<-<-<-<-<-<-<-<-<-<-<-<-<-
    /                                 /
   /                                 /
  /                                 /
 /                                 /
->->->->->->->->->->->->->->->->->-

At no point in this pattern is he in a sideslip, or doing something weird. You could do this pattern as well when there's no wind at all, the only difference is that in that case it follows an usual groundtrack. There's just one reason to follow this weird pattern today: because with the wind acting on it, this pattern will be bent into a nice rectangular groundtrack, which is what Bob wants.
And there's no real danger here, except when Bob does stupid things (like flying close to a stall and then entering a turn). The only danger you could argue there is that two of his turns are somewhat longer than the usual 90 degrees.
Related, you seem to think that crabbing is somehow difficult or dangerous. But whether you are crabbing or not makes no difference at all as long as you're not close to the ground. Only landing with a crab is difficult, because you have turn the nose (de-crab) or your landing gear will push you the wrong way.
In a sense, a plane flying due north while over the equator is crabbing heavily, because its nose points due north while its "groundtrack" as seen from space is east-northeast. Because it's moving along the Earth's rotation of ~1,000 mph. But does this matter? No, not as long as you don't want to interact with any satellites or other things outside the Earth's atmosphere.

I think the last paragraph, while technically correct, is confusing and should be removed.

If you made no correction for the wind on crosswind and base, the pattern shape would actually resemble a trapezoid.  And yeah, that last paragraph is a mess.  Not remotely related to the question, and who would use space as a frame of reference for an aircraft in the atmosphere?!

We might as well view the track of the plane even further out from interstellar space   https://youtu.be/0jHsq36_NTU

Crosswind, headwind, tailwind - in whichever direction, it only answers to Einstein in that all airspeed is relative, and relative airspeed is the only relevant contribution to lift generation. All forces and responses are relative to aircraft axes, as also described in this answer. Wind relative to the aeroplane, it can only see headwind when flying!

That is, wind at constant velocity and direction. Start making sudden changes, and now there are effects to lift and drag. A sudden horizontal wind gust exerts an additional force and causes an acceleration, direction of which depends on the direction of the gust. The magnitude of the acceleration depends on the magnitude and duration of the gust, and on the inertia of the plane.
The questions about sudden downwind turns are based on a partial realisation of the above gustiness effect. The thing is, a gust in longitudinal air flow creates an almost instantaneous effect due to the inertia of the aircraft. A manoeuvre on the other hand can never be instantaneous due to again the inertia. In aircraft of several tons and designed and optimised for their mission, manoeuvres are changes that can be considered as being quasi-static. Sudden downwind turns can simply not be sudden enough.
For a bumble bee: perhaps it can spin instantaneously so that its backside points into its flight velocity, but I don't think so.

Going to have to try it in the wind tunnel.  Take a wing, offset it to the windstream by 10-20 degrees, and roll it down, then up. Graph lift, derive AOA changes from sidewind component striking the top, then the bottom of the wing.  But good to hear the stories and knowledge of the others with experience. Hoping PK will weigh in on this one.

Worth noting: this would be testing roll during sideslip.

@RobertDiGiovanni You do not bank your plane to compensate for the wind. You just fly a different heading. Your question is then like: when I turn from heading 010 (or 350) to 090, do I risk a stall more than when turning from 000 to 090?

@Suma logical, but wrong.  You can't "fly a different heading" when the runway points in one direction.  Take time to learn the techniques of holding a ground track. That is what this is about.

Off course you can and it is done all the time. It is just before the touchdown you change from a crab to a sideslip. See http://www.flaps2approach.com/journal/2014/6/17/crosswind-landing-techniques-part-one-crab-and-sideslip.html : "Just before flare, left rudder will be applied to correct for drift to bring aircraft into line with centerline of runway.  This technique is called 'de-crabbing’. "

@RobertDiGiovanni - Quick question. In one statement, you said that you landed a C172 by crabbing into the wind with wings level (I presume since that would be a proper crab). Then, you performed a side slip right before touchdown to align yourself with the runway. While this is one way to approach the runway for landing (the preferred way), the other method would be to hold a side slip on the entire approach. The first way would give you a Heading that is different from the runway heading although your ground track same . The second way would give you a heading & track the same as the runway.

My question is - was this in an actual 172 in real life, in real time? Or, was this virtual? If it were in a real C172, how were you performing your approaches up till now? And, if this is a real C172, do you have a glass cockpit with heading, course, and track indicators? Or, do you at least have serviceable steam gauges with heading bugs on a DG/HSI with little to no drift and a good OBS? The reason I ask is with good gauges, you would see the difference between your heading and course be increasingly different the stronger the wind is in a crabbing situation.

Anecdotally, I have crabbed into the wind on a practice ILS IAP with the wind near my demonstrated crosswind limit for a low wing Archer. My course and ground track were 179°. My wings were level. My slip/skid indicator was centered. Yet my heading was about 150°. I was able to hold that coordinated crab with minor rudder input to combat weathervaning since I was at or below 1500 RPM. Once past the flare and in the round out, I lowered the upwind wing and pushed opposite rudder to maintain ground track while aligning with the runway. Touched down on the upwind main first.

@Dean F. eyes on runway serve as heading indicators (steam gauges?), but I can see from the cross country definition of "wings level crab" with power, a power-off rudder/aileron into wind may seem radically different.  We all have our ways.  My experience was real.  But, as we have gone delightfully off subject (this is a study of the downwind turn myth) I can tell you I approached at 65 knots, no power, used a coordinated rudder/aileron to hold track, rounded, then used sideslip/rudder to land.  Using sideslip/rudder from say short final was also taught to me and tried on occasion.

But now I have to wonder if "wings level crab" is more fuel efficient than leaning over a bit, but some may swear by it for the sake of passenger comfort.  This isn't rocket science, it's better!

@Suma rudder is applied to point the nose down the runway, ailerons are used to "correct for drift" (laterally), since the crosswind tries to push the plane off the centerline.

I apologize if my questions seem impertinent. I am just curious as to what formed your perspective and opinion on the current subject.

Thank you for your answer. What has shaped my perspective is about equal time between glass & steam gauges. In a Cessna 172 with steam gauges, to hold track in a rectangular pattern in a stiff wind, wings level crab worked as well as it does on approach. In those cases, I could not tell you exactly the crab angle. It was noticeably more than 10°. You could especially tell during low approaches down a 7000 foot runway where flying uncoordinated with one wing low would have been a hazardous situation. Keeping the wings level and flying the crab was the key.

Each 90° turn in the pattern was done with the wind and crab angle in mind. Meaning half the turns were more than 90° heading and the other half were less than 90° by about the same difference. He ground track remained rectangular. And the flight coordinated.

Now, I have flown a cross-wind approach completely without crabbing to have one more thing in my toolbox. We side slipped for the wind  (not a front slip to lose altitude) the entire way down. We maintained a stable approach although this was uncoordinated flight.

Since I fly with a Stratus that can be linked to ForeFlight, CloudAhoy, or Google Earth, I have the data to show my ground track as well as, now, my heading and attitude with the linked AHRS.

@RobertDiGiovanni - Why are you doing your approaches with no power like you suggested? Are you practicing emergency procedures? Are you doing power out 180’s? Are you starting are Final leg consistently too high? Or, is this your technique for a standard approach? The textbook approach calls for reduced power, not no power. You may want to start doing it the textbook way. Doing it your way is going to make instrument approaches to Checkride standard a lot harder to maintain. Pitch for speed, and Power for altitude (on the backside of the power curve), as they say.


text:
Is V best endurance configuration the best way to save a heavily overloaded aircraft?
After reading this, we all agree "the experienced pilot takes over, lowers the nose, and the plane is saved".
Not bad advice, but what speed offers the greatest amount of excess thrust required to climb (not excess "power").  Is Lift/Drag ratio really highest at Vbg, or is it actually highest at Vmin sink?  Would increased prop efficiency actually make an airspeed closer to Vx with flaps a better way to prevent a crash?
We all have learned that lift is less than weight in a climb, but could increasing Coefficient of Lift with flaps and flying Vx be the best chance to save the day?

Related - https://aviation.stackexchange.com/questions/8365/what-is-the-difference-between-best-rate-of-climb-and-maximum-rate-of-climb/8368#8368 , https://aviation.stackexchange.com/questions/55355/why-is-best-angle-rate-of-climb-indicated-in-airspeed/55371 , https://aviation.stackexchange.com/questions/73997/how-can-best-glide-speed-be-lower-than-best-rate-of-climb-speed/74346#74346 (really good answer, and contradicts my comment above, for some cases), https://aviation.stackexchange.com/a/74004/34686 (similar),

More - https://aviation.stackexchange.com/questions/21350/under-what-conditions-can-the-maximum-angle-of-climb-be-achieved-for-jet-and-pro , https://aviation.stackexchange.com/questions/420/how-can-i-calculate-maximum-rate-of-climb/12727#12727 , https://aviation.stackexchange.com/questions/53081/in-a-jet-are-vy-and-vx-achieved-at-a-constant-aoa-at-all-altitudes/53185#53185 , https://aviation.stackexchange.com/questions/77391/why-isnt-vy-lower-than-the-speeds-for-best-l-d-ratio-and-best-glide-ratio-in-an (question includes links to several more related q's)

@quietflyer So, we are looking a Vx.  Please take out your previous comments and give others a chance.  (But Vx is higher at higher weight, and Vx/Vy convergence is a very good point).

I guess the really scarey part is being able to lift off but not being able to climb out of GE.  This is probably where "zero hours" began tugging on the yoke. Observations of the 1903 telemetry at Kitty Hawk leads to the same conclusions, and that Wilbur made 300 yards that 1st day by not trying too hard to climb.

1903 telemetry?!

Typically aircraft have a specific speed "best L/D" which is different than the other speeds cited, and it will increase as weight increases. The curve would be specific to the aircraft and almost certainly would be with flaps retracted.

@theMayer  this was the type of answer I was looking for.  Vx and Vy converge at absolute ceiling (so you don't have to go that high, just find the V at a lower power setting).  The convergence seems to be near Vbg, a little toward Vx, because of  better prop thrust output at lower speeds.  A V absolute ceiling formula would be neat.  It would apply to "V lifting most weight".  Slat/flap configuration would be dependent on type of aircraft.

It seems one could argue that all the speeds converge at the aircraft's max ceiling, and an overloaded plane that can barely climb is essentially at its max ceiling.  So the question should perhaps be how does the speed that keeps you air in this situation, compare (in terms of IAS?) to the speeds for Vx, Vy, Max Range, Max Glide Angle, etc that would be valid at the same density altitude for a more reasonable weight, like the max allowable gross weight under those conditions?  Just brainstorming a little here, might offer some insight into a way to sharpen the focus of the question--

@quietflyer  the ounce of prevention is far better, eh.  I think the gist is the effect of maximum thrust output speed relative to lowest drag (Vbg).

@RobertDiGiovanni Unfortunately, I think the engine is a confounding factor here. What are we trying to optimize?

@theMayer  we are trying to optimize the airspeed that provides maximum lift.  It appears to be between Vx and Vbg, but, I agree that will be weight dependent (up to absolute ceiling for that weight).

@RobertDiGiovanni if just looking at theory (which applies to properly loaded airplane), I'm confused as to why we are looking at an actually overloaded airplane to frame the question?

@theMayer to help save it

Re "It seems one could argue that all the speeds converge at the aircraft's max ceiling" -- well, I guess I meant all the speeds pertaining to power-on operation.  Not V best-glide-angle.

Unfortunately, I don't think this question is answerable, at least not with a definitive answer, because the answer is that it is undefined.
In the linked video, we're clearly looking at an aircraft that is being operated well outside of its design limits (overweight, out of balance, possibly high density altitude). Talking about which of the V_sub numbers is best to fly in this case is absurd, because all of those numbers assume the aircraft is being operated within its design performance envelope.
I have attached a basic L/D chart below. Notice how the lines behave toward the edges. A plane being operated outside of the envelope will be off the left side of this chart, where the numbers increase asymptotically. What this is saying is that there is eventually a configuration where the plane is incapable of flying at ANY number. I think the plane in the video was perilously close to that threshold.
As a glider pilot, we have the idea of "best speed to fly," which may be minimum sink, maximum glide (L/D), or any speed in the operating envelope of the aircraft. Min sink is usually just above stall speed, but you're moving pretty slow. It's ideal for climbing in a thermal, where you want to move slow. However, if you're doing a glider race, you'll want ballast in the wings for extra boost to the optimal l/d speed, so you can travel farther at a faster speed.
If you're in an area of sinking air, you'll go faster than best L/D. Flaps add drag. We only want to extend flaps when we're looking to decrease stall speed and fly slowly, at the expense of needing additional power to fly at that low speed. But, the only guidance for an overloaded plane is to remove weight before trying to take off.

Re:  your comment:  V best endurance is a powered aircraft.  We can agree Vbg is best speed aerodynamicly, but excess thrust is higher at lower speeds.  That is why V best endurance may actually provide max lift.  With the 172, Vx is flown with flaps 10.  Your answer may be applicable to jets.

You'd probably need to be traveling at jet takeoff speed to get enough lift for that overloaded plane... best endurance is not going to be shaped like this, and won't have the aerodynamics taken into account. If your airplane will stall at anything lower than full throttle, you're stuck at full throttle and best endurance speed is the same.

That speed is published in the POH for the aircraft. For a 172, it's 75-85 kias, and I haven't looked, but there's probably a chart to go along with it. Irrelevant if you're unable to achieve that airspeed due to excessive induced drag.


text:
Could catalytic combustion improve the efficiency of jet engines?
Catalytic converters have come into common use with the millions of automobiles on the road these days, greatly reducing their emissions of unburnt fuel and NOx.  Because the converter is outside the engine, the considerable amounts of heat they produce is waste.
After learning catalytic converters are also being used on indoor propane heaters and even wood stoves, could they be installed on jet
flame cans?
Platinum/palladium catalytic heaters require only sufficient fuel vapor and a certain amount of heat to complete combustion of hydrocarbons and other byproducts to CO2, H2O, and N2.
As seen with woodstoves and propane heaters, the amount of fuel usage is significantly less than flame alone (which makes much more shorter wavelength visible light).
For example, per BTU, a catalytic heater running on gasoline (fumes) burns around 30% less fuel (observed by the writer) than even a ceramic plate propane heater (with a constantly lit pilot flame).
With the jet, added heat from the converter should increase thrust per pound of fuel burned.  I dare say catalytic afterburners might be awesome.
Has any of this been tried?

I can't see a mechanism where you could use the heat from the catalytic reaction to create extra thrust.

@GdD  Catalytic heaters give more complete combustion.  Reference "jet exhaust" on Google (CO, NOx, soot, particulate matter).  A catalytic surface enables these flame byproducts to be converted more completely to CO2, H2O, and N2.  In addition to cleaner air, these reactions are exothermic.  Ask any mechanic how hot a CC can get.  The converter would be inside the engine, adding heat along with the flame can.

https://aviation.meta.stackexchange.com/a/3493/20174

Catalytic converters, once warmed up, work so well that the flame can be extinguished and the converter alone will consume all of the fuel.  A jet engine burns around 1 liter per second.

As an addendum to Anonymous Physicist's answer:
Car type catalytic converters consist of mesh or honeycomb structures seeking to get the majority of the molecules in the stream exposed to the catalyst. By definition this prevents optimal supersonic flow since there must be interaction between fixed parts of the cat structure and the entire flow volume, reducing the available thrust.
This paper doing CFD on a car exhaust seems to be focusing on flow rates under 10 meters a second, vs several hundred for a jet engine, and seems to indicate that at that speed the needed length was 2 meters to allow reactions to complete. It would seem likely operating at increased flow velocity would require a longer catalyst bed.
Taken together this means that adding a catalytic converter to a jet engine it would involve massively increasing the flow area at or just downstream of the combustion area by a factor of between 10-100 reduce the skin drag by lowering speed while in the converter and keep overall system length sane, it would then have to constrict again to get useful final exhaust velocity which has performance impact as well.
The overall high frontal area shape from this would be problematic for a aircraft engine but might do something for a stationary power plant, assuming there is actually much partial combustion going on. Reciprocating engines as used in cars have partial combustion since there are areas of the cylinder where either fuel/air mix is not correct or where the thermal mass of the structure inhibits combustion. A gas turbine burner section does not need to perform the other three elements of a combustion cycle (compression, expansion and exhaust) so can optimise both mixing and temperature to achieve high combustion efficiency.
As an alternative to a catalytic converter bed, just lengthening the combustion area would appear to have a similar result of maximizing combustion completion and it is notable that power turbines like the LM2500 do not appear to have longer combustion zones compared to their aviation parents suggesting that the performance costs of messing with the high velocity flow pushes increase in combustion completion into diminishing returns territory.

Better answer.  What is the flow rate in the flame can?  Is it anywhere close to supersonic?  "Overall high frontal shape" was never a problem for centrifugal compressors, or, indeed fans.  We need some hard data on conditions in the flame can, removing NOx alone would be a very positive step forward.  I'm not buying "thrust loss" quite yet, as pressure drop begins at the turbine.

@RobertDiGiovanni - some envelope math suggested a cat chamber 10 meters wide by 20 long, need some big gains for that to be worth while but also do not trust the numbers. I think the bigger problem is in the last para - I could not find any mention of unburned products in jet exhaust when I went looking at papers, the current focus seems to be NOx control by avoiding hot spots.

That might definitely help commercial flights in our "good" ozone layer.

Ignoring drag losses, the force generated by a jet engine is (mdot fuel + mdot oxidant)*exhaust velocity. (mdot is mass flow rate: mass per time)
If the catalyst retards the velocity of the exhaust jet, then you are just killing the force generated by the jet engine.  And when I think of a catalytic converter, I think of long, narrow channels and lots of pressure drop.
Also, since the active surface of the catalyst is only doing the job of catalyzing the oxidation, the larger you make the engine, the worse the pressure drop will be per volume of gas.  This is a classic surface area to volume scaling gotcha.

Pressure drop is at the turbine, not in the flame can.

@RobertDiGiovanni Wait 'til you put that catalytic converter in.

agree that an auto style may significantly impede flow.  The combustors on wood stoves look more like round waffles.  I have seen them scrub a propane flame, and (once fully warmed) handle all of the propane flow (with the flame out).  Consider that the catalytic rxn is faster than flame.  So,  a platinum/tungsten woven mesh at the end of the flame can would best Test # 1.

How is the catalytic rxn faster than a "flame"?  I would like to see some data.  Flames only need to mix and react.  While the catalyst may increase the rate of reaction while it is on the catalyst, overall it is slowed by the time for adsorbing onto the catalyst, desorbing from the catalyst, and that's after the reactants have to diffuse through the materials surrounding the catalyst in the first place.  Since you want to use the "waffles" which have a relatively low surface area, you are not going to burn everything on the catalyst.

"flames only need to mix and react" at a much higher temperature.  The dwell time for combustion is very short and the mixture (because of turbulence) is not consistent within the flame.  Platinum is on the surface and very fast.  See 3rd comment.

@RobertDiGiovanni  I requested something substantive.  Show me some rate data, and explain how the high temperature catalyst doesn't ignite the unreacted gas near the hot catalyst.

Actually, I'm the one asking the question:  "has any of this been tried ".  I have seen "waffle" combustors work with no flame using wood smoke or propane.  Kerosene will be vapor in the fuel can.  To answer your query, my point was that flame requires higher temperature.

@RobertDiGiovanni Actually, you asked a question about "catalytic combustion" improve the efficiency of jet engines.  You then make grandiose claims about how "catalytic combustion" works, and then "poo poo" classic issues with kinetic theory and reactor design.  Oh, well.

OK, let's do a .1 meter flame with an air mass flow rate of 5 meter per second.  What's the dwell time in the flame?  1 second/5 meters x 0.1 meters = 0.02 seconds.  Just please try a little platinum and call me in the morning.

A fluid catalyst would not make any sense.  It would get ejected from the engine, defeating the point.
A solid catalyst could improve efficiency, but the improvement would be minuscule.  Solid catalysts work at the surface.  Compared to the mass flow through a jet engine, the surface available to be coated with catalyst inside the engine is extremely small.  This ensures poor catalyst performance.  Considering that catalysts add weight and require periodic replacement of precious metals, this is not an economically plausible idea for aviation.
You cannot increase the surface area of the jet engine or its exhaust nozzle because there would be a large drag penalty.  Jet engine shapes are highly optimized.
This might make sense if you were using the jet engine as a snow melter on the ground.  That's a rare situation.  Usually it is better to move the snow instead of melting it.

Just to help a little, the cat(s) would be in the high pressure, lower flow section, on or near the flame cans.

I think the answer made it quite clear in paragraph 3 why you cannot use a powder or honeycomb structure in a jet engine.

If most of the mass flow is not involved in combustion because it's nitrogen or excess oxygen, then the catalyst won't work on it.

"the surface available to be coated with catalyst inside the engine is extremely small."  Yes well the surface area inside a typical automatic exhaust pipe is quite small too, but they still manage to make it work by using a honeycomb structure (e.g. https://s.hdnux.com/photos/77/25/50/16603680/3/rawImage.jpg). You could do the same in a jet engine, but there would be a big weight penalty, which is a bigger deal for plane than a car, and also there would be a pressure drop across the structure, which would hurt performance.  I don't know whether it would be a net win or not.

@RobertDiGiovanni do you have any maths for the increased drag/lost thrust from sticking a high area cat matrix in the exhuast stream? it looks like for cars adding the catalytic converter is around 5% performance loss due flow restriction, but cars are not depending on exhaust flow rate for thrust.

@RobertDiGiovanni you probably couldn't put the catalyst right near the "flame cans" (I would call them fuel nozzles) as the temperature would be too high and would degrade performance (or even melt the catalyst). I think you'd need to be downstream of the high pressure turbine, (typically a much "cooler" ~2000F versus the ~3000F turbine inlet temperature)

@DanielK valid point about temperature melting the converter.  That's why metals (tungsten, titanium, ceramics) used to make the flame can could be incorporated into the converter.  A platinum/tungsten alloy mesh may be possible.  "Downstream of the turbine" may work, but recovery of energy by finishing combustion is what I'm after.


text:
Is Vbg maximum Lift/Drag ratio, or is it really maximum Speed/Drag ratio?
A question about "minimum power" airspeed had me wondering, why not think of "maximum Lift/Drag ratio" as maximum Lift/Thrust ratio.
Gliders would be included, as they use gravity for propulsion.
It seems that the condition of maintaining linear flight using the least amount of fuel (altitude) per unit time is V min sink.
So what is Vbg?  It seems to be the maximum Speed/Drag ratio, but it burns more fuel per unit time than V min sink.

Power = Thrust x Airspeed

Would it be more accurate to describe Vbg as maximum Speed/Drag or Power/Drag (Thrust = Drag at steady state) ratio?
Looking at the units for Lift and Drag uncovers something troubling:

Lift is a Force = ma = mv/s that is constant regardless of speed.

Drag = Thrust = mv/s varies with speed and Angle of Attack

"Glider thrust" = fuel burn = mg×delta h/s = mv$^2$/s.

You see, the units of lift and the units of fuel burn do not cancel to form a proper ratio (no units).
Lift × Speed = mv/s × v = mv$^2$/s.  Now one can divide this by fuel burn to get a meaningful ratio.
So "Lift/Drag" appears to be a bit of a curiosity$^1$.  Here, slower may be better.
$^1$ analysis of the Clark Y airfoil show similar L/D ratio between AOA of +3 and +7 degrees.

From a glider perspective, min sink is generally slower than best glide; in the former case a certain amount of forward speed is sacrificed to give a relatively small improvement in sink rate.  With a glider, the energy that goes into maintaining flight comes from a loss of height (in still air) and so power is directly proportional to sink rate and so min sink is the point at which power is at its minimum, by definition.  I think the essence of your question relates to the difference between energy per unit time vs energy per unit distance travelled

Does this answer your question? Why is the L/D ratio numerically equal to the glide ratio?

I think this answer covers the distinction quite well: Why would maximum fly-time (endurance) not coincide with max L/D operating point?

Not a duplicate- I'm pretty sure we've never before had a question about maximizing the quantity Speed/Drag.

"Best glide" in what sense? Time or distance? Assuming still air, or a tail/head wind.

@MaxPower distance, and, as your mentioned in comment, the horizontal component (earth referenced).

@RobertDiGiovanni, I'm trying to understand what it is you're looking for exactly.
For V minimum sink (i.e. staying in the air as long as possible, Maximum endurance), we don't care about the speed (we're not going anywhere) and hence we try to minimize drag for a given Lift equal to Weight.
For the V best glide (i.e. getting as far as we can, max range), staying in the air alone is not enough, we need to minimize the power used, so we need to maximize drag times V for a given Lift equal to Weight.
These goals are the same regardless of where the energy (initially) comes from.

@ROIMaison sort of stumbled my way to an easy "pocket proof" the L/D max is at Vbg.  Vmin sink had me confused, but using Force vectors (with gravity) w/out V (power) has provided a solution. Thanks.  See answer w/ diagrams.

@ROIMaison -- re "Maximum endurance), we don't care about the speed (we're not going anywhere) and hence we try to minimize drag for a given Lift equal to Weight." -- this is not actually true.  See my answer, including my comments below (which I'll eventually integrate into answer)-- Drag is actually minimized at the speed for best L/D, not the speed for min sink.  Min sink happens to be the speed where we minimize the quantity ((lift coefficient cubed)/(drag coefficient squared)).  And best glide doesn't minimize Drag * V, it simply minimizes Drag.

This answer is for gliders, or airplanes with failed engines or engines producing zero thrust.  To adapt for a powered airplane with operating engine, in a descending glide under partial power, simply substitute "Drag-Thrust" for "Drag", throughout the answer.
The speed for best glide ratio (in still air) is the speed for maximum L/D.
For reasonably flat glide angles it is easy to show that Lift is only slightly less than Weight, and nearly all the variation in the L/D ratio, as airspeed is varied, is due to changes in Drag.  A graph of L/D versus airspeed will have a rather broad "plateau" where we can vary the airspeed above or below the optimum for best L/D, without making a huge change in L/D.  By the logic above, we are also not making a huge change in D itself.  Therefore it's very clear that if we are trying to maximize Airspeed/Drag, this will certainly come at a higher speed than the speed for maximum 1/D (i.e. minimum Drag), which is very nearly the same as the speed for maximum L/D.
Therefore the speed for best glide ratio (in still air), which is also the speed for maximum L/D, and is also very nearly the speed for maximum 1/D,  cannot also be the speed for maximum Speed/Drag. The speed for best glide ratio (in still air) is always slower than the speed for maximum Speed/Drag.
(It's not obvious that the quantity "Speed/Drag" should have any particular practical significance to a glider pilot, or to a power pilot for that matter, but that need not prevent us from talking about it.)
PS as you read this answer, please don't confuse Lift and Drag with their respective coefficients.  While it happens to be true that the ratio of L/D is also the ratio of the Lift coefficient to the Drag coefficient, in most applications we can't simply interchange Lift and Drag with their respective coefficients, and most of the statements about Drag in this answer would not also be true of the Drag coefficient.

If one grinds through some numbers maximum (ratio of) speed to a given amount of drag will be similar maximum L/D, too slow and induced (from AoA) is less efficient, too fast and form drag is too much.  It somehow made more sense to equate drag to "Glider thrust".  But you are right, speed is not the critical factor for distance, glide angle is.

Are yo using lift in the sense of normal to the surface of earth/gravity or normal to the fuselage? Most uses I see are normal to the fuselage. A force vector could be broken into into horizontal and vertical components(relative to earth), or parallel to flight path and perpendicular to the flight path.

@MaxPower  - the accepted definition of Lift is well established.  See related ASE questions about L/D ratio and glide ratio.  Maybe I should add links to some of them, to this answer. PS in gliding flight Lift is not defined as normal to the earth, and also is not defined as normal to the fuselage.

@MaxPower -- see for example https://aviation.stackexchange.com/a/81831/34686

This answer needs some more improvement-- it states that the speed for max L/D is not exactly the same as the speed for max (1/D) (i.e. the speed for min D).  In fact you can see from the L-D-W vector triangle that these two speeds are identical.  I've actually already stated that in a previous answer-- "The vector diagrams above demonstrate that as long as Weight is fixed, when we maximize L/D, we also minimize Drag."-- https://aviation.stackexchange.com/a/81831/34686.

"A graph of L/D versus airspeed will have a rather broad "plateau" where we can vary the airspeed above or below the optimum for best L/D, without making a huge change in L/D. "-- even if the graph had a very "peaked" shape, the speed that maximizes Speed/Drag will still always be higher than the speed that maximizes 1/Drag, which is the same as the speed that maximizes Lift/Drag.

@quietflyer Exactly my point, it is defined dependent on context. I should have said relative to bulk airflow rather than fuselage, that was sloppy. If lift is defined as vertical (relative  to earth) then wing lift of a glider will never be exactly equal to weight because there is also a vertical component of the drag vector. While in airfoil design lift is always normal to the bulk air flow; in this case total lift is the vector which gives a vertical component equal to weight minus the vertical component of drag, which could mean total lift is greater than weight.

@MaxPower total lift is greater than weight it would not be steady state.  The vertical lift vector triangle is closed by the component of the gravity vector pulling the plane forward.  Drag simply is opposite the direction of flight.  Reduction in Lift requirement is weight - (cosine angle of descent x weight).  Factors involved in increase of drag at Vmin sink are:  increased AoA, increased form drag, increased tail drag, greater tail downforce.  At Vmin sink drag slows the plane down, even though it's angle of descent is steeper.

By that logic flying in a circle cannot be steady state because the total lift is greater than the weight of the aircraft (Approximately 1.4*weight  in a turn created by a 45degree bank to achieve a vertical component of lift equal to the weight.)  Gravity pulls down and only down, not forward, it is the same with a sled sliding down a slope, gravity provides the energy and vertical force but it is the hill that applies the horizontal component of force. Likewise the wing's lift acts like the hill providing a horizontal forward force to counter drag.

@MaxPower hear you there, one can consider drag against the angled bottom of the glider as the driving force.  That is one way of seeing it.  Yes, gravity only pulls "down", but from the plane reference it also provides a forward force component.  I particularly like it because sin angle × W provides the "sum of all drags" in a glide.

@RobertDiGiovanni Gravity is not providing any forward force in any reference. The forward force only comes from the lift vector being slightly forward. "Weight" is created by lift not by gravity. Gravity isn't really a force at all, the "force" is the ground or lift constantly accelerating you up to maintain a relative position. A weird effect of curved spacetime theoretically linked to the forward flow of time. When "falling" [in a vacuum] you are stationary in spacetime but spacetime its self is moving toward the earth...

... like being in a lifeboat, you really don't move relative to the water but you are going along with the general ocean current so you are moving relative to the stuff which is resisting the flow (and thus encountering a force). But with gravity the water is spacetime its self. This also helps explain time dilation in gravity wells and why GPS time is not equal to earth surface time, if the dilation was to accumulate it would cause gps nav error of 6miles per day. Clearly its not a perfect analogy, one is a 2D surface and the other is 4D.

@MaxPower  describing reality in mathematical terms and formulas can be local.  In other words, "earth is flat" is good enough within a few km, but certainly not in orbit.  In aviation, gravity is a force of constant magnitude and direction.  When an aircraft has a vertical component towards earth, it has an acceleration = G × sin angle of descent × weight.

@MaxPower. Draw the vectors, make it a 90 degree descent, gravity pulls forward, any lift curves the aircraft flight path.  I was a big believer in the "forward lift vector" originally too.

@MaxPower - re gravity not providing any forward force- depends on how define forward- food for thought : see https://aviation.stackexchange.com/a/56371/34686  and https://aviation.stackexchange.com/a/75480/34686

You are correct I should have said horizontal force not "forward", I was using forward in the broad sense of horizontal travel rather than local motion, the difference between a glider and a stone. In local relative motion there is no "forward" lift vector because lift by definition is a force perpendicular to flow direction. But there can be a horizontal lift vector if there is any non-zero vertical component of motion. Gravity still only has a vertical component and therefore cannot provide any horizontal force, but gravity can provide vertical motion and thus allow some horizontal lift.

One way to resolve the issue of units is to consider "glider thrust" to be the sin of the angle of descent × the weight vector.  Now there are all force units as mv/s or simply pounds.
It then becomes clear that, allowing for minor differences in lift requirements for angles of descent, modeling follows reality in that:

Best speed for maximum distance glide, after satisfying the lift requirement, will occur at Vbg.

A steeper angle of descent is used at V min sink (the plane does not glide as far).  This resolves as higher AoA and lower speed keep one in the air longer, but more drag is used for lift, and less on airspeed.

Lift is similar, gravitational "push" is greater, therefor L/D is less.

A steeper angle of descent at lower AoA produces the same result: a shorter glide path, will more drag than Vbg from greater speed.

sin angle of descent x weight translates as "Drag"

What is "best speed for distance"? Are you trying to maximize Airspeed / Sink rate?  If so, that quantity is essentially the definition of the glide ratio or very nearly so, for reasonably flat glide angles.  (Wouldn't work for a falling brick!)

Minimum sink speed corresponds to minimum power required, and thus, to best endurance speed, the ratio best range speed/best range speed is 1.317. More info here: https://aviation.stackexchange.com/questions/36440/why-is-the-ratio-maximum-range-speed-maximum-endurance-speed-the-same-for-any-ai

Just trying to rationize it with consistant units.  Essentially, the flattest glide angle makes most efficient use of gravity.


text:
Is the amount of dihedral on low wing aircraft based more on landing considerations than on aerodynamic benefits?
During a recent discussion on glider design, a quandary arose over what the aerodynamic effects of moving an otherwise unchanged wing from the top of an aircraft to the bottom would be.
One claim was that it would reduce "self righting" characteristics due to a change in the interaction of the relative wind to the upwind wing/fuselage root and reduced effect on the lift of the downwind wing.
This would be the equivalent of increased "anhedral effect".
Another point of view was that lowering the wing lowers CG, increasing area above the CG, which (as an upright tail fin or fuselage component) would increase tendency to roll away from a side wind, or increased "dihedral effect".
In the mean time, high wings and low wings continue to fly on in abundance, with passionate arguments supporting the virtues of each.
With so much ambiguity, even after more than 120 years of aviation to sort things out, is it possible that both designs still exist because one really does not have a significant aerodynamic advantage over the other?
Landing considerations of long winged aircraft are obvious considering the dangers of catching a wing tip and "cartwheeling". Increased dihedral and a high wing position help prevent this.  Others state low wings are easier maintenance.
In the air, are they really closer than we think?

The two models pictured where thrown from a 25 foot hill in a mild crosswind.  They landed so close to each other they nearly collided.

Many related questions could be linked, including those referencing the "pendulum effect", which most of us understand isn't actually very analogous to the physics at play in a real pendulum, yet the vertical placement of the CG relative to various surfaces of the aircraft clearly does have some effect on roll stability, due to effects of surface area above/below CG in a sideslip... anyway to start that list of linked questions-- https://aviation.stackexchange.com/q/12099/34686, https://aviation.stackexchange.com/q/26396/34686, https://aviation.stackexchange.com/a/703/34686,

Here's a point which may often go under-appreciated in such discussions-- imagine a wing with no dihedral or anhedral, and imagine that we can consider that the wing has no "side area".  It still makes a contribution to the total drag of the aircraft.  If the wing is above or below the CG, then during a sideslip, the orientation of that drag vector is such that it does create a rolling moment around the aircraft's longitudinal axis, even if no actual "sideforce" is generated (i.e. no force perpendicular to the flight path in the direction perpendicular to the lift vector), only L and D.

(ctd) So even in a pure "flying wing" design w/ no fuselage/wing interaction, or fuselage side area, or actual anhedral or dihedral, if we mount a heavy weight on a long skinny pole protruding high above or below the wing, we'd still create some anhedral-like or dihedral-like effect. The point being, just because a wing has minimal "side area", and no actual dihedral or anhedral, doesn't mean that it's only contribution to the balance of roll torques during a sideslip is its influence on the vertical location of the CG, by virtue of its mass.

A possible related new ASE question could be posed as to whether, in case of fuel tanks positioned well below or above aircraft CG (think typical high-wing or low-wing GA planes with fuel tanks in the inboard portions of the wings), status of tanks full versus empty has a noticeable effect on aircraft's roll stability...

@quietflyer The fact that it has not been definitively answered makes me wonder, remember the leeward wing is blanked to some extent.  I think, from your glider experience, a very long, thin wing may be more towards anhedralling underneath.  Re airliners, is large amounts of dihedral (to avoid wing tip strikes but promoting Dutch Roll) actually a design compromise?  Let's please condense comments down to one.  Thanks!  PS look what I had to do to the "high wing" tail to get it to turn into the wind!

Likely will get moved to chat soon :; but that's all from me for now...

Dihedral cannot be chosen in isolation. For good flying characteristics the parameter it controls, namely $\text{c}_{l\beta}$, must have a certain relation to weathervane stability $\text{c}_{n\beta}$. While $\text{c}_{n\beta}$ is positive, $\text{c}_{l\beta}$ is negative, so if you plot both as in the figure below (taken shamelessly from Ray Whitford's Fundamentals of Fighter Design), the ratio can be found in the second quadrant of a coordinate system.

If weathervane stability is high, dihedral must also be higher than usual so the ratio of both coefficients falls into the "satisfactory" area. If the engine-out case drives tail size up, the wing dihedral has to follow so that flying characteristics do not suffer.
The hasty development of the Heinkel 162 resulted in too much dihedral and adverse flying qualities. The quick fix was to add Lippischohren (Lippisch ears), named after the person who used them first, Alexander Lippisch.

He-162 front view (picture source). No, these are not inverted winglets but Lippischohren to  correct excessive dihedral.

That's probably where Canadair got the idea for the CL415 tips, for precisely the opposite purpose.

See Grumman F9F Panther as a follow-on.

@JohnK Same for the horizontal tail of the MPP version of the Diamond DA42.

Some points to consider:

Dihedral effect requires sideslip to work. When an airplane banks due to a disturbance, you must have some sideslip develop to create the self-righting differential lift in the first place.  The vertical tail volume must be low enough to allow some sideslip to develop.  If the fin/rudder is too large, sideslip instantly generates a weathervaning yaw and you get a spiral diving tendency instead of an automatic return to wings level tendency.  So there is a whole balancing act on fin sizing you have to deal with first.
Pendulum effect to the extent that it exists, isn't a pendulum so much, rather than a form of static stability in a vertical plane (you're floating in a gas - a pendulum requires a fixed attachment).  If the aerodynamic center of the body in the vertical axis, the vertical Neutral Point, is offset from the roll rotation axis, the vertical C of G, then when a lateral movement is present the Neutral Point will seek to trail, or weathervane, behind the center of mass so to speak, just as in the yaw and pitch axis. If the Neutral Point is above the vertical C of G, a restorative moment is created similar to dihedral effect because the body with the wind acting on it wants to "trail" the vertical C of G. Take a Cessna 172, and add a heavy weight on the end of a long rigid pole below the fuselage, and skid the plane, and it will roll away from the skid from static stability forces acting in the vertical axis (it's not the "pendulum effect" of the weight down below, it's the body of the plane trying to weathervane behind the lowered center of mass, about which the airplane is rolling, and in fact the same effect will work in weightlessness). I would say that although this effect is present to some degree in most airplanes, slightly positive in high wing airplanes and close to neutral in low wing ones, it's too subtle to notice because the vertical Neutral Point and center of mass are too close together in most cases.  However this effect can be observed on float planes due to the mass of the floats moving the vertical C of G down more than they move the aerodynamic center down, and in my experience airplanes have somewhat noticeably better roll stability when on floats than when on wheels.
Then there's dihedral effect from actual geometric dihedral of the wings, and dihedral effect that high wing airplanes enjoy from the T configuration, where sideslip generates a righting force just from the lack of side surface area above the wing. A low wing airplane will need more wing physical dihedral than a high wing one to have the same righting effect during sideslip.  Lots of high wing airplanes do fine with no wing dihedral at all, just relying on the T junction of wing-to-fuselage. The Wittman Tailwind homebuilt and the CL215/415 water bomber come to mind, designs where mild dihedral effect was all that was wanted, so the wings have zero geometric dihedral. In fact, when the CL415 was developed from the 215, one of the nasty surprises was that the flat PW120 engine nacelles sticking up above the wing killed off some of the T junction effect, as if it had become a mid-wing airplane, and roll stability was insufficient.  The odd turned up wingtips of the CL415 are there to create some geometric dihedral effect to compensate, without having to redesign the entire wing (one of the ultimate aerodynamic bandaids).

So in the end, a low wing airplane with a dihedral angle of, say 3 degrees, might have the same roll stability, that is, righting force from sideslip, as a high wing airplane with a T configuration plus 1.5 degrees of geometric dihedral.  In the end it's what do you have to do to achieve a given righting tendency from sideslip with whatever configuration you're using, and if you moved a wing from the top to the bottom, you would expect to have to increase (or introduce) dihedral angle to maintain the same righting tendency due to sideslip.
As far as landing conditions go, not really. The best example is the Corsair, where anhedral is used to shorten the gear legs, but must be offset by major dihedral outboard to achieve the desired roll stability.  If the Corsair's wings were made straight with longer gear legs, the overall dihedral angle would be way less than the outboard gull wing's has to be.

I'm also wondering if:  back in the pre-FADEC days was "snaking" from variations in engine thrust adding to the design woes more than today?

@RobertDiGiovanni  Yeah you look at a C172 there is a bit of dihedral, looking to be between 1 and 2 deg, but less than a low wing Piper which is upwards of 4 or 5 or something.  On engines, once an engine is set it's  pretty constant FADEC or no, so I've never heard of that being a problem. What FADEC gets you is the ability to operate much closer to the margins and limits, which is where most of the efficiency and responsiveness benefits come from, plus a bit  ease of use benefit by automating some of the of power setting, etc. Note that pre FADEC engines are still electronically trimmed.

Fun experiment.  Make the vertical tail surface of your glider massively oversize and see what it does.  It should develop a tendency to turn on its own a lot more, maybe entering a spiral every time you throw it.

Done that with paper airplanes (BB before balsa).  Absolutely, and excessive anhedral makes it worse.  I had high hopes for upright big fins at first (yawing into the wind and accelerating the leeward wing), but found that what happened in a crosswind (with dihedral) was the plane would roll away from the wind, sink, and turn downwind, especially when I went from 30 cm wings to 1 meter wings (these were free flight, no controls allowed).  Getting it to roll into the wind really helps it turn into the wind, hence a (Predator-like) downward facing fin.

For some configurations (significantly swept wing with tail-mounted engines) anhedral absolutely can be a design concern for takeoff/landing. For example, on Tu-134 you can have wingtips strike before you get tailstrike. And with minimum roll, on Tu-154.

But in that case the anhedral was still for stability and control purposes, to reduce the effect of sweep + high wing, and the tip strike problem was an "undesirable consequence".  The question is about dihedral angles specifically selected for landing gear considerations, like the Corsair.  Only one other airplane I can think of fits that category, the Avro Arrow interceptor, where the primary purpose of the delta wing's anhedral was to shorten the gear legs.

I'm not sure about such narrow interpretation of the question. It seems generic enough. Re Tu-134 et al, it is conceivable (i.e.: speculation) that even greater anhedral would be [aero]dynamically beneficial, but it couldn't be done due to "landing considerations".


text:
Does parasitic drag increase exponentially with velocity?
A deep dive into the Cessna 172 POH showed a remarkable linearity between rate of fuel consumption and velocity at various cruise power settings ranging from 55 to 75%.  Originally a study of propeller efficiency, it lead to an observation of the drag equation:

Drag = $\rho$ × Area × Cdrag × V$^2$

Power has the units kg V$^2$/time.  Fuel as potential energy has the units kg V$^2$, and it's use rate as kg V$^2$/time.
Power is also expressed as Force × distance/time or kg V/s x V.
At steady state thrust force = drag force.
Examination of the drag equation $\rho$×AxCd×V$^2$ seems at first to imply Power is proportional to V$^3$.
Yet, when the units of the drag equation were written out, it was found that, in order to yield force kg velocity/time there could be only one V!
Running through the units revealed that "kg" must come from kg/m$^3$ $\rho$ × m$^2$ Area × m distance, leaving V/s as the remainder of the Drag force kg V/s.
The implications are that parasitic drag increases linearly with velocity, rather than exponentially, as often depicted,
and that drag in terms of V$^2$ is actually an expression of Force × Distance or Work.
Consideration of density kg/m$^3$ x Volume of Air m$^3$ (as Area m$^2$ x distance m) has me wondering is the drag force equation being misunderstood as a Work equation?
A possible source of confusion is that, at airspeed "behind the power curve" (slow), induced drag surely makes the drag curve look exponential.  But the Cessna data makes the fuel consumption kgV$^2$/s perfectly proportioned between drag kgV/s and airspeed V at cruise speeds.  Not V$^3$!
Many people sing songs of praise for cruise climb, and indeed, lowering Angle of Attack and using greater speed seems to help fuel economy.  What's going on here?
I will gladly migrate to physics if no one here would like to try to explain this.

Comments are not for extended discussion; this conversation has been moved to chat.

You've messed up your units. $\rho$ has units of $\rm kg/m^3$, area has units of $\rm m^2$, and velocity has units of $\rm m/s$. Putting this all together:
$$ \left[\rm Drag\right] = \rm\frac{kg}{m^3}\cdot m^2 \cdot \left(\frac{m}{s}\right)^2 = \frac{kg\,m}{s^2} $$
which has dimensions of force, and the drag equation containing $v^2$ is dimensionally consistent. Side note: $v^2$ scaling is called "quadratic," "exponential" refers $2^v$ scaling.
Regarding cruise climb, higher climb speeds do not improve fuel efficiency on a per hour basis. Either way, the engine is at full power so it doesn't really change at all. But since you are going faster, you go farther in the same amount of time, so they can improve fuel efficiency on a per mile basis.

@RobertDiGiovanni Thrust has units of $\rm kg,m/s^2$. Power has units of $\rm kg,m^2/s^3$. You get from thrust to power by multiplying by velocity, not velocity squared.

@RobertDiGiovanni Drag is also a force, and the power required to overcome drag is $v$ times the drag, which scales as $v^3$, not $v^2$. You can't relate these directly to fuel consumption numbers since the propeller efficiency is a function of airspeed and $C_d$ depends on the angle of attack and thus depends indirectly on airspeed, as well as directly to some degree depending on flow conditions.

Aerodynamic forces can be expressed in dynamic pressure $q$ and a reference area of the aeroplane (usually the wing area $S$). In order to account for the shape of the plane, dimensionless coefficients are used, C$_D$ for drag and C$_L$ for lift. So drag =  $C_D \cdot q \cdot S$ and lift = $C_L \cdot q \cdot S$.
Parasitic drag is a component of the total drag, and is therefore according to the above definition a linear function of dynamic pressure $q$. And for subsonic flow, Bernoulli has found that $q$ = ½ $\rho$ V$^2$.

Does parasitic drag increase exponentially with velocity?

No. For incompressible flow, parasitic drag increases quadratically with velocity.
That's. It.

V2/V1 has unit 1. It seems to me that you’re lost in an improper unit analysis. Converting fuel potential energy into net kinetic power has many efficiency losses which can be a function of airspeed, but that has no consequences for the units, which are correct.

Drag is a force. Thrust is a force as well. Forces are what moves objects, which will reach a constant velocity if all resulting forces are zero.

Seeing as how both lift and drag vary as the square of the airspeed, it would surprise me enormously if parasitic drag, (except for small edge case effects), was any different. An exponential relationship would certainly be out of the question.


text:
Why did the 777-9 leave landing gear down during first flight?
In Sam Chui's video covering the first 777-9 flight, there's a shot of the plane over the clouds with landing gear still down.

I thought maybe the shot was from another flight. I couldn't find another video with exactly that shot, but CNET's video shows the landing gear remaining down for a long time after takeoff.
Assuming I'm correct, that the gear remained down for the whole flight: why?

I'm fairly confident Boeing does this for all first flights as a safety precaution, but I will let someone with more direct knowledge provide an official answer